Relevant equations:

$\displaystyle \Delta K = W$

$\displaystyle K = \frac{1}{2}mv^2$

$\displaystyle W = F*d$

$\displaystyle F_f = \mu F_N$

Determine the truck's change in kinetic energy. Note that the final kinetic energy will be zero because it has no final velocity.

$\displaystyle K_i = \frac{1}{2}(3500kg)(25\frac{m}{s})^2=1,093,750J$

$\displaystyle K_f=0$

$\displaystyle \Delta K = K_f - K_i = -1,093,750 J$

Using the work-energy theorem, set change in kinetic energy equal to the work done by friction.

$\displaystyle \Delta K = -F_f d$

Substitute the equation for force of friction.

$\displaystyle \Delta K=-\mu F_Nd$

$\displaystyle \Delta K = -\mu (mg)d$

Use the given values for coefficient of friction, mass, and acceleration of gravity to solve for the distance.

$\displaystyle -1,093,750 J= -(0.4)(3500kg)(9.8\frac{m}{s^2})d$

$\displaystyle d = 79.7m \approx 80m$

The two horizontal forces acting on the box are the applied force of 40N to the right, and the friction force ($\displaystyle \mu_kF_N$)to the left. So the net horizontal force is written as the equation below.

$\displaystyle 40N-\mu_kF_N=ma$

$\displaystyle F_N=mg$ is the normal force. Solving for acceleration, a, we get the equation below.

$\displaystyle a=\frac{40N-\mu_kF_N}{m}$

We know the following from the question.

$\displaystyle m=14kg$

$\displaystyle g=9.8\frac{m}{s^2}$

$\displaystyle \mu_k=0.2$

$\displaystyle F_N=mg=(14kg)(9.8\frac{m}{s^2})=137.2N$

Thus, the acceleration is

$\displaystyle a=\frac{40N-(0.2)(137.2N)}{14kg}$

$\displaystyle a=0.9\frac{m}{s^2}$

Relevant equations: 

$\displaystyle F_{net}= ma$

$\displaystyle F_f=\mu F_N$

$\displaystyle F_g = mg$

Write net force equations for horizontal and vertical directions:

Horizontal: $\displaystyle F_{net, x} = F_{pull, x}-F_f$

Vertical: $\displaystyle F_{net,y}=F_N+F_{pull,y}-F_g$

Since the crate is not accelerating, $\displaystyle \small a=0\frac{m}{s^2}$ in each direction, implying that $\displaystyle \small F_{net,x}=F_{net,y}=0N$.

Horizontal: $\displaystyle 0 = F_{pull,x}-F_{f}$

Vertical: $\displaystyle 0 = F_N + F_{pull,y}-F_g$

Express the horizontal and vertical components of the pulling force in terms of the total pulling force, $\displaystyle F_{pull}$.

Horizontal: $\displaystyle 0 = (F_{pull}*cos(45^o))-F_f$

Vertical: $\displaystyle 0 = F_N + (F_{pull}*sin(45^o))-F_g$

Replace force of friction with its expression in terms of the normal force.

Horizontal: $\displaystyle 0 = (F_{pull}*cos(45^o))-(\mu F_N)$

Rearrange to isolate the normal force.

Horizontal: $\displaystyle F_N =\frac{F_{pull}*cos(45^o)}{\mu}$

Substitute this term for the normal force in the net vertical force equation.

Vertical: $\displaystyle 0 = (\frac{F_{pull}*cos(45^o)}{\mu})+(F_{pull}*sin(45^o))-F_g$

Solve to isolate the pulling force.

$\displaystyle F_{pull}=\frac{F_g*\mu}{cos(45^o)+\mu*sin(45^o)}$

Use the given values for the mass of the box, coefficient of friction, and acceleration of gravity to solve for the pulling force.

$\displaystyle F_{pull}=\frac{(15kg)(9.8 \frac{m}{s^2})(0.20)}{cos(45^o)+0.20*sin(45^o)}$

$\displaystyle F_{pull}=\frac{29.4N}{0.849}=34.65N\approx35N$

We can calculate the work done by friction by using $\displaystyle W=F_{friction}\cdot d$.

It is just the product of the friction force and the length of the ramp. Rewrite work as $\displaystyle W=\mu_kF_Nd$.

Since this problem involves a box sliding down an incline, the normal force is $\displaystyle F_N=mg\cos\theta$.

Work is now written as the equation below.

$\displaystyle W=\mu_kmg(\cos\theta) d$

We know the following information from the question.

$\displaystyle \mu_k=0.4$

$\displaystyle m=287kg$

$\displaystyle g=9.8\frac{m}{s^2}$

$\displaystyle \theta=33^{o}$

$\displaystyle d=5.5m$

using these values, we can calculate work from our equation.

$\displaystyle W=(0.4)(287kg)(9.8\frac{m}{s^2})(\cos(33))(5.5)$

$\displaystyle W=5189 J$

First, we know that the tension on the string is $\displaystyle 40\:N$ at an angle of $\displaystyle 30$ degrees from the horizontal. We must find the x and y components of the tension force.

$\displaystyle T_y = T\sin(30) = 40\cdot\sin(30)= 20\:N$

$\displaystyle T_x = T\cos(30) = 40\cos(30)= 34.64\:N$

Next, we must find the normal force acting on the block. The normal force of the block itself is $\displaystyle 15\:kg \cdot 9.8\:m/s^2 = 147\:N$, but there is a string pulling on it. We must subtract the y component of tension to get the final normal force

$\displaystyle F_n = 147-20 = 127$

Next, we know that the block is being pulled to the right, and the frictional force is acting in the opposite direction. So we get the equation:

$\displaystyle T_x - F_f = 0, F_f = \mu F_n$

$\displaystyle 34.64 - \mu 127 = 0$

$\displaystyle \mu = 0.273$

$\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a))\\&F_{friction}=51kg(9.81\frac{m}{s^2})(sin(11^{\circ})\\&F_{friction}=95.46N\end{align*}$

$\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{friction}=100kg(9.81\frac{m}{s^2})(sin(73^{\circ}))\\&F_{friction}=938.13N\end{align*}$

$\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{friction}=19kg(9.81\frac{m}{s^2})(sin(69^{\circ}))\\&F_{friction}=174.01N\end{align*}$

$\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{friction}=(F_{mass}-F_{rope})*sin(a)\\&F_{rope}=F_{mass}-\frac{F_{friction}}{sin(a)}\\&F_{rope}=2786.04N-\frac{207.04N}{sin(48^{\circ})}\\&F_{rope}=2507.44N\end{align*}$

$\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{friction}=(F_{mass}-F_{rope})*sin(a)\\&F_{friction}=(101kg(9.81\frac{m}{s^2})-693.57N)(sin(68^{\circ}))\\&F_{friction}=275.6N\end{align*}$